1. Class Greeting

2. Parallel Lines and Proportional Parts
Chapter 7 – Lesson 4
Parallel Lines and Proportional Parts

3. Objective: The students will be able to solve problems by using the Parts of Similar Triangles.

4. What is…?
An Altitude?
A median?
An angle bisector? B A C

5. p. 544

6. K In the figure, is an altitude of and is an altitude of Find x if and
Answer: JI = 28.
Example 5-3a

7. In the figure, ΔABC ~ ΔFGH. Find the value of x.
Answer: x = 14

8. In the figure, ΔLJK ~ ΔSQR. Find the value of x.
MK and TR are corresponding medians and LJ and SQ are corresponding sides. JL = 2x and QS = 2(5) or 10.
~Δ have corr. medians proportional to the corr. sides.
Substitution
12 ● 10 = 8 ● 2x Cross Products Property 120 = 16x Simplify = x Divide each side by 16.
Answer: x = 7.5

9. In the figure, is an altitude of and is an altitude of Find x if and
Answer: 17.5
Example 5-3c

10. The side opposite the bisected angle is now in the same proportion as the sides of the angle.

11. Find x.
Since the segment is an angle bisector of the triangle, the Angle Bisector Theorem can be used to write a proportion..
Triangle Angle Bisector Theorem
9x = (15)(6) Cross Products Property
9x = 90 Simplify.
x = 10 Divide each side by 9.
Answer: x = 10

12. Find n.
Answer: 15

13. Cross Products Property
Find PS and SR.
∆  Bisector Theorem
Substitution
40x – 80 = 32x + 160
Cross Products Property
Subtraction
8x – 80 = 160
8x = 240
Addition
Division
x = 30
Substitution
PS = 30 – 2
SR =
PS = 28
SR = 25
Simplify

14. by the ∆  Bisector Theorem.
Find AC and DC.
by the ∆  Bisector Theorem.
Substitute in given values.
4y = 4.5y – 9
Cross Products Theorem
–0.5y = –9
Simplify.
y = 18
Divide both sides by –0.5.
Answer: DC = 9 and AC = 16.

15. Kahoot!

16. Lesson Summary:
Objective: The students will be able to solve problems by using the Parts of Similar Triangles.

17. Preview of the Next Lesson:
Objective: The students will review for Lesson 7-5 to 7-6 test.

18. Stand Up Please